Question

Решите неравенство:
log1/3 (x-2)+log1/3 (12-x)>2

Answer 1

$log_{1/3}(x-2)+log_{1/3}(12-x) textgreater 2; ,; ; ; ODZ:; left { {{x textgreater 2} atop {12 textgreater x}} right. ; ,; 2 textless x textless 12\\log_{1/3}(x-2)(12-x) textgreater log_{1/3}(1/3)^2\\-x^2+14x-24 textless 1/9; |cdot (-9)\\9x^2-126x+216 textgreater -1\\9x^2-126x+217 textgreater 0\\frac{D}{4}=(frac{126}{2})^2-9cdot 217=2016=16cdot 126; ,; sqrt{frac{D}{4}}=4sqrt{126}\\x_1= frac{63-4sqrt{126}}{9} approx 2,01; ;; x_2= frac{63+4sqrt{126}}{9} approx 53,95\\xin (-infty , frac{63-4sqrt{126}}{9} )cup ( frac{63+4sqrt{126}}{9} ,+infty )$

$2 textless x textless 12quad Rightarrow \\xin (2;; frac{63-4sqrt{126}}{9} )$